Problem: The differentiable functions $x$ and $y$ are related by the following equation: $y=2x^2-x$ Also, $\dfrac{dx}{dt}=-0.1$. Find $\dfrac{dy}{dt}$ when $x=-3$.
Explanation: Let's start by differentiating the equation $y=2x^2-x$ with respect to $t$. $\begin{aligned} y&=2x^2-x \\\\ \dfrac{dy}{dt}&=4x\cdot\dfrac{dx}{dt}-1\cdot\dfrac{dx}{dt} \\\\ \dfrac{dy}{dt}&=(4x-1)\cdot\dfrac{dx}{dt} \end{aligned}$ We are given that $\dfrac{dx}{dt}=-0.1$, and we want to find $\dfrac{dy}{dt}$ when $x=-3$. Let's plug ${x=-3}$ and ${\dfrac{dx}{dt}=-0.1}$ into the equation we obtained: $\begin{aligned} \dfrac{dy}{dt}&=(4{x}-1)\cdot{\dfrac{dx}{dt}} \\\\ &=(4({-3})-1)({-0.1}) \\\\ &=1.3 \end{aligned}$ In conclusion, when $x=-3$, the value of $\dfrac{dy}{dt}$ is $1.3$.